Using Faraday's Law to calculate the magnetic field

[Blanchdog]

Homework Statement

Consider an electric field E = E0 cos(k⋅r - ωt +φ), where k is orthogonal to E0, r is a position vector, and φ is a constant phase. Show that B = (k X E0)/ω * cos(k⋅r - ωt +φ) according to Faraday's Law.

Relevant Equations

Faraday's Law: ∇ X E = -∂B/∂t

I got stuck near the beginning, so I tried working backwards. Starting from

B = (k X E0)/ω * cos(k⋅r - ωt +φ)

I found

-∂B/∂t = -k X E0 sin(k⋅r - ωt +φ)

So now I need to find ∇ X (E0 cos(k⋅r - ωt +φ)) and see that it is equal to the above result. This is where I'm stuck though, I'm not sure how to take the curl of this electric field because of that dot product of k and r, leaving the field as a scalar (as far as I can tell). Help is much appreciated!

 

[Delta2]

Homework Statement:: Consider an electric field E = E0 cos(k⋅r - ωt +φ), where k is orthogonal to E0, r is a position vector, and φ is a constant phase. Show that B = (k X E0)/ω * cos(k⋅r - ωt +φ) according to Faraday's Law.
Relevant Equations:: Faraday's Law: ∇ X E = -∂B/∂t

I'm not sure how to take the curl of this electric field because of that dot product of k and r, leaving the field as a scalar (as far as I can tell)

Scalar is only the cosine term. The $\mathbf{E_0}$ is a vector though it is constant. You have to use the following vector calculus identity: $$\nabla\times (\mathbf{E_0}f)=(\nabla\times\mathbf{E_0})f+\nabla f\times\mathbf{E_0}$$ where $f$ is your scalar that is $$f=\cos(\mathbf{k}\cdot\mathbf{r}-\omega t+\phi)$$
Also because $\mathbf{E_0}$ is a constant vector we have $\nabla\times\mathbf{E_0}=0$ which simplifies further the above identity.
And as I said , scalar is only the cosine term that is $f$, the product $\mathbf{E_0}f$ is a vector so we are justified to take its curl.

 

[berkeman]

Hi @Blanchdog -- As you can see from the reply by @Delta2 it makes the math much more readable if you use LaTeX to post equations. That is the gold standard at PF.

Please look through the LaTeX Guide link at the bottom left of the Edit window to become familiar with the basics of posting in LaTeX. Thank you.

 

[Blanchdog]

Scalar is only the cosine term. The $\mathbf{E_0}$ is a vector though it is constant. You have to use the following vector calculus identity: $$\nabla\times (\mathbf{E_0}f)=(\nabla\times\mathbf{E_0})f+\nabla f\times\mathbf{E_0}$$ where $f$ is your scalar that is $$f=\cos(\mathbf{k}\cdot\mathbf{r}-\omega t+\phi)$$
Also because $\mathbf{E_0}$ is a constant vector we have $\nabla\times\mathbf{E_0}=0$ which simplifies further the above identity.
And as I said , scalar is only the cosine term that is $f$, the product $\mathbf{E_0}f$ is a vector so we are justified to take its curl.

Thanks Delta2! With that vector identity I was able to prove the relation.